hdu 4746Mophues[莫比乌斯反演]

Mophues

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 327670/327670 K (Java/Others)

Total Submission(s): 1669 Accepted Submission(s): 675

　



Problem Description

As we know, any positive integer C ( C >= 2 ) can be written as the multiply of
some prime numbers:

C = p1×p2× p3× ... × pk

which p1, p2 ... pk are all prime numbers.For example, if C = 24, then:

24 = 2 × 2 × 2 × 3

here, p1 = p2 = p3 = 2, p4 = 3, k = 4

Given two integers P and C. if k<=P( k is the number of C's prime factors), we
call C a lucky number of P.

Now, XXX needs to count the number of pairs (a, b), which 1<=a<=n , 1<=b<=m, and
gcd(a,b) is a lucky number of a given P ( "gcd" means "greatest common
divisor").

Please note that we define 1 as lucky number of any non-negative integers
because 1 has no prime factor.



Input

The first line of input is an integer Q meaning that there are Q test cases.

Then Q lines follow, each line is a test case and each test case contains three
non-negative numbers: n, m and P (n, m, P <= 5×105.
Q <=5000).



Output

For each test case, print the number of pairs (a, b), which 1<=a<=n , 1<=b<=m,
and gcd(a,b) is a lucky number of P.



Sample Input

2

10 10 0

10 10 1



Sample Output

63

93



Source

2013 ACM/ICPC Asia Regional Hangzhou Online



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//Source：http://acm.hdu.edu.cn/showproblem.php?pid=4746


Description(题意):

任何整数C
( C >= 2 )都可以写成素数之积

C = p1×p2×
p3×
... × pk

其中， p1, p2 ... pk 是素数。如
C = 24, 则
24 = 2 ×
2 × 2
× 3,
其中, p1 = p2 = p3 = 2, p4 = 3, k = 4.

给定两整数 P和 C,
若 k<=P ( k是
C的素因子个数),称
C是P的幸运数.

现小X需计算的点对 (a,
b)的个数,其中1<=a<=n
, 1<=b<=m, gcd(a,b)是 P的幸运数
( “gcd”是最大公因数).

注意：因为1无素因子，定义1为任何非负数的幸运数.

Input

首行有一个整数
T，表示有 T 组测试数据.接下来有T行，每行是一种测试数据，含3个非负整数n,
m 与P (n, m, P <= 5×105.
T <=5000).

Output

对每种测试数据，输出对
(a, b)的个数，其中 1<=a<=n , 1<=b<=m,
且 gcd(a,b)
是 P的幸运数.

Sample
Input

2

10 10 0

10 10 1

Sample
Output

63

93

//num[j]记录j的因子数。
//g[j][num[i]]用于计算具有相同个数的素因子的i的?(j/i)之和，
#include<cstdio>
#include<iostream>
using namespace std;
typedef long long ll;
const int M=5e5+5,N=19;
int n,m,p,T,g[M]
,num[M];
int tot,prime[M/3],mu[M];bool check[M];
int calc(int y,int x){
int res=0;
while(!(y%x)) y/=x,res++;
return res;
}
void sieve(){
n=5e5;mu[1]=1;
for(int i=2;i<=n;i++){
if(!check[i]) prime[++tot]=i,mu[i]=-1;
for(int j=1;j<=tot&&i*prime[j]<=n;j++){
check[i*prime[j]]=1;
if(!(i%prime[j])){mu[i*prime[j]]=0;break;}
else mu[i*prime[j]]=-mu[i];
}
}
for(int i=2;i<=n;i++) if(!num[i]) for(int j=i;j<=n;j+=i) num[j]+=calc(j,i);
for(int i=1;i<=n;i++) for(int j=i;j<=n;j+=i) g[j][num[i]]+=mu[j/i];
for(int i=1;i<=n;i++) for(int j=1;j<19;j++) g[i][j]+=g[i][j-1];
for(int i=1;i<=n;i++) for(int j=0;j<19;j++) g[i][j]+=g[i-1][j];
}
ll solve(int n,int m,int p){
if(p>=19) return 1LL*n*m;
if(n>m) swap(n,m);
ll ans=0;
for(int i=1,pos=0;i<=n;i=pos+1){
pos=min(n/(n/i),m/(m/i));
ans+=1LL*(n/i)*(m/i)*(g[pos][p]-g[i-1][p]);
}
return ans;
}
int main(){
sieve();
for(scanf("%d",&T);T--;){
scanf("%d%d%d",&n,&m,&p),
printf("%I64d\n",solve(n,m,p));
}
return 0;
}